The equation x2+ ax + 2023 = 0 has roots x1 = b + 1 and x2 = 8b2 + 1 for some real number b. Find a + b.
So far I have been trying Vieta's and getting 8b3 + 8b2 + b +1 = 2023, -a = 8b2 + b + 2. And then getting 2023 + a = 8b^3, but then I don't know what to do from here, is my approach totally wrong? Also, can I have the answer with a solution, so I understand how you got the answer?
The sum of the roots of a quadratic equation is given by -b/a, and the product of the roots is given by c/a, where a, b, and c are the coefficients of the quadratic equation.
In this case, we know that the sum of the roots is b + 8b^2 + 2, and the product of the roots is 2023.
We can use these values to find a and b.
(b+1)+(8b^2+1)=b+8b^2+2 2b+2=b+8b^2+2 b=8b^2
We know that b is a real number, so b cannot be 0. Therefore, b = 1/8.
We can now find a.
-b/a+c/a=b+8b^2+2 -1/(8a)+2023/a=b+8b^2+2 -1/(8a)+2023/a=1/8+2023/8 2022/(8a)=2025/8 a=2025/2022
Therefore, a + b =2025/2022+1/8 =2051/2022.
